$‖⋅‖_p(1≤p≤∞)$在$ℝ^n$(有限维)等价,但在$ℓ^1$不等价

hbghlyj

对任何$\mathbf x\inℝ^n$,定义
\[\|\mathbf x\|_p\coloneqq\left(\sum_{i=1}^n|x_i|^p\right)^{\frac 1 p}\]
MSE - Are all the $ℓ^p$ metrics topologically equivalent for positive integer values of $p$?
由这帖知,$‖⋅‖_p(1≤p<∞)$与$‖⋅‖_∞$等价,因为
\[\lVert x \rVert_\infty \leq \lVert x \rVert_p \leq n^{\frac 1 p}\lVert x \rVert_\infty\tag1\]
对任意$\mathbf x∈ℝ^n$成立.

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范数的等价有传递性, 可以得出$‖⋅‖_p(1≤p<∞)$与$‖⋅‖_q(1≤q<∞)$等价.
具体来说, 在“范数的等价有传递性”的证明中, 利用(1)导出了不等式:
\[\lVert x \rVert_p \leq n^{\frac 1 p}\lVert x \rVert_\infty\le n^{\frac 1 p}\lVert x\rVert_q\tag2\]
与
\[n^{-\frac 1 q}\lVert x \rVert_q \leq \lVert x \rVert_\infty\le \lVert x\rVert_p\tag3\]
因此,对$1\le p,q<∞$,存在常数$c,C$使得
$$c\| \mathbf x \|_q \leq \| \mathbf x \|_p \leq C\| \mathbf x \|_q$$
对任意$\mathbf x∈ℝ^n$成立.(由(2),$c=n^{-\frac1q}$满足不等式.由(3),$C=n^{\frac1p}$满足不等式.)

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问题:对固定的$p,q$,求$c,C$的最佳值.
$c$的最佳值是1. 对于$\mathbf x=(1,0,0,⋯,0)$可以取等. 又叫 范数不等式.
$C$的最佳值是$n^{\frac1p-\frac1q}$. 对于$\mathbf x=(1,1,1,⋯,1)$可以取等. 又叫 幂平均不等式.

hbghlyj

在$ℓ^1$上,$‖⋅‖_p$都有定义(因为它们都$≤‖⋅‖_1$,所以是有限的).
$‖\mathbf x‖_∞≤‖\mathbf x‖_2$, 所以$B_∞({\bf x},r)⊃B_2({\bf x},r)$.
但是在$ℓ^1$上, $‖⋅‖_∞$与$‖⋅‖_2$不等价.
令${\bf 0}=(0,0,…)$. 假设存在$r>0$使$B_2({\bf0},1)⊃B_∞({\bf0},r)$.
设$0< a< r$和${\bf a}_k=(\underbrace{a,a,…,a}_k,0,0,…)$.
则${‖{\bf a}_k‖}_∞=a⇒{\bf a}_k∈B_∞({\bf0},r)⇒{‖{\bf a}_k‖}_2=a\sqrt k$
$a\sqrt k$当$k\to\infty$可以任意大(明确来说,${\bf a}_k∉B_2({\bf0},1)$ 对 $k>a^{-2}$), 与$B_2({\bf0},1)⊃B_∞({\bf0},r)$矛盾.

hbghlyj

Wilson A. Sutherland - Introduction to Metric and Topological Spaces-Oxford University Press (2009) page 70-71
Example 6.36 On the set $\mathcal{C}[a, b]$ of all continuous real-valued functions on $[a, b]$ the sup metric $d_{\infty}$ and the $L^1$ metric $d_1$ are not topologically equivalent. To see this, let $f, g \in \mathcal{C}[a, b]$. Then $|f(t)-g(t)| \leqslant d_{\infty}(f, g)$ for all $t \in[a, b]$ by definition of $d_{\infty}$. Hence by integration theory,
\[
d_1(f, g)=\int_a^b|f(t)-g(t)| \mathrm{d} t \leqslant(b-a) d_{\infty}(f, g),
\]
which is 'half of a Lipschitz equivalence' and gives $B_{\varepsilon}^{d_{\infty}}(f) \subseteq B_{(b-a) \varepsilon}^{d_1}(f)$ for any $f \in \mathcal{C}[a, b]$ and any $\varepsilon>0$.

However, if we let 0 denote also the constant function with value 0 , then $B_1^{d_{\infty}}(0)$, which is $d_{\infty}$-open, is not $d_1$-open. For if it were then we would have $B_{\varepsilon}^{d_1}(0) \subseteq B_1^{d_{\infty}}(0)$ for some $\varepsilon>0$. But for any $\varepsilon>0$ there exists a continuous function $f$ on $[a, b]$ such that $d_1(f, 0)<\varepsilon$ yet $\|f\|_\infty=1\Rightarrow f \notin B_1^{d_{\infty}}(0)$ (see Figure 6.1 for the graph of such a function).

hbghlyj

hbghlyj 发表于 2022-10-23 19:05
Wilson A. Sutherland - Introduction to Metric and Topological Spaces-Oxford University Press (2009)
...
which is 'half of a Lipschitz equivalence' and gives $B_{\varepsilon}^{d_{\infty}}(f) \subseteq B_{(b-a) \varepsilon}^{d_1}(f)$ for any $f \in \mathcal{C}[a, b]$ and any $\varepsilon>0$.
...

说明在$\mathcal{C}[a, b]$上, $d_∞$导出的拓扑比$d_1$导出的拓扑 更精细(finer topology)

hbghlyj

hbghlyj 发表于 2022-10-23 17:15
...
$‖\mathbf x‖_∞≤‖\mathbf x‖_2$, 所以$B_∞({\bf x},r)⊃B_2({\bf x},r)$.
但是在$ℓ^1$上, $‖⋅‖_∞$与$‖⋅‖_2$不等价.
...

说明在$\ell^1$上, $d_2$导出的拓扑比$d_∞$导出的拓扑 更精细

hbghlyj

S.6.pdf
Proposition S.6.12 The norms $\|\cdot\|_1$ and $\|\cdot\|_{\infty}$ are not Lipschitz equivalent on $\mathbf{l}^1$; neither are the norms $\|\cdot\|_2$ and $\|\cdot\|_{\infty}$ on $\mathbf{l}^2$. The norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are not Lipschitz equivalent on $\mathbf{l}^1$.
Proof To prove the first assertion, it is enough to show that there is no positive constant $k$ such that $\|\boldsymbol{x}\|_1 \leqslant k\|\boldsymbol{x}\|_{\infty}$ for all $\boldsymbol{x} \in \mathbf{l}^1$. But given any $k>0$ we can choose an integer $n>k$ and let $\boldsymbol{x}=(1,1, \ldots, 1,0,0, \ldots)$ with $n$ non-zero entries. Then $\boldsymbol{x} \in \mathbf{l}^1$ and $\|\boldsymbol{x}\|_{\infty}=1$ while $\|\boldsymbol{x}\|_1=n>k=k\|\boldsymbol{x}\|_{\infty}$.
An entirely similar proof shows that $\|\cdot\|_2$ and $\|\cdot\|_{\infty}$ are not Lipschitz equivalent on $1^2$.
Finally, consider $\boldsymbol{x}^{(\boldsymbol{n})}=(1,2,3, \ldots, n, 0,0, \ldots)$. Then $\boldsymbol{x}^{(\boldsymbol{n})} \in \mathbf{l}^1$. Also, from elementary formulas we have $\left\|\boldsymbol{x}^{(\boldsymbol{n})}\right\|_1=n(n+1) / 2,\left\|\boldsymbol{x}^{(n)}\right\|_2=\sqrt{n(n+1)(2 n+1) / 6} \leqslant(n+1)^{3 / 2}$. Now given any $k>0$ we may choose an integer $n$ such that $k(n+1)^{3 / 2}<n(n+1) / 2$, since the power of $n$ on the left is less than on the right. This gives $k\left\|\boldsymbol{x}^{(n)}\right\|_2<\left\|\boldsymbol{x}^{(n)}\right\|_1$. Hence there is no constant $k>0$ such that $\|\boldsymbol{x}\|_1 \leqslant k\|\boldsymbol{x}\|_2$ for all $\boldsymbol{x} \in \mathbf{l}^1$. This shows that $\|\cdot\|_1$ and $\|\cdot\|_2$ are not Lipschitz equivalent on $\mathbf{l}^1$.

hbghlyj

Relations between p norms

Recall Hölder's inequality
$$
\sum\limits_{i=1}^n |a_i||b_i|\leq
\left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}
$$
Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$
$$
\sum\limits_{i=1}^n |x_i|^p=
\sum\limits_{i=1}^n |x_i|^p\cdot 1\leq
\left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}}
\left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}
$$
Then
$$
\Vert x\Vert_p=
\left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq
\left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}=
\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\=
n^{1/p-1/q}\Vert x\Vert_q
$$
In fact $C=n^{1/p-1/q}$ is the best possible constant.
3. For infinite dimensional case such inequality doesn't hold. For explanation see this answer